ExamJEE Classes

Solving Algebraic Limits (Form 2)

$latex \displaystyle \infty -\infty \quad Form &s=1$

Indeterminate forms problems are simplified (generally rationalised ) first there after they generally acquire $latex \displaystyle \left( \frac { \infty }{ \infty } \right) \quad Form &s=1$

For Example

$latex \displaystyle=\lim _{ x\longrightarrow \infty }{ \left( x-\sqrt { { x }^{ 2 }+x } \right) } \\ =\lim _{ x\longrightarrow \infty }{ \frac { \left( x-\sqrt { { x }^{ 2 }+x } \right) }{ 1 } } \times \frac { \left( x+\sqrt { { x }^{ 2 }+x } \right) }{ \left( x+\sqrt { { x }^{ 2 }+x } \right) } \\ =\lim _{ x\longrightarrow \infty }{ \frac { \left( x-\sqrt { { x }^{ 2 }+x } \right) }{ \left( x+\sqrt { { x }^{ 2 }+x } \right) } } \\ =\lim _{ x\longrightarrow \infty }{ \frac { -x }{ \left( x+\sqrt { { x }^{ 2 }+x } \right) } } \\ =\lim _{ x\longrightarrow \infty }{ \frac { -x }{ x\left\{ 1+\sqrt { 1+\frac { 1 }{ x } } \right\} } } \\ =\lim _{ x\longrightarrow \infty }{ \frac { -1 }{ 1+\sqrt { 1+\frac { 1 }{ x } } } } \\ =-\frac { 1 }{ 2 } \quad \left[ as\frac { 1 }{ x } \longrightarrow 0\quad as\quad x\longrightarrow \infty \right] &s=1 $

 

April 18, 2019
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